Question
For each subset S of R given below, determine whether S is bounded above,
bounded below, bounded, or neither. If S is bounded above, determine
sup(S), and decide whether or not sup(S) is an element of S. If S is bounded
below, determine inf(S), and decide whether or not inf(S) is an element of
S.
1. S =
n
1
n
o
| n ∈ Z − {0} ;
2. S = {2x | x ∈ Z};
3. S = [−1, 1] ∪ {5} = {x ∈ R | − 1 ≤ x ≤ 1} ∪ {5};
4. S =
n
x
2y
5. S =
n
n+1
n
n
o
| x, y ∈ N ;
o
|n∈N ;
³
6. S = (−1)n 1 +
1
n
´
o
|n∈N ;
7. S = {x ∈ Q | x2 < 10};
8. S = {x ∈ R | |x| > 2};
Answer
1. Bounded above by 1 (since for n ∈ Z − {0}, either n ≥ 1 in which case
1
≤ 1, or n ≤ −1, in which case n1 ≤ 0), and so has a supremum. Since
n
1 is an upper bound for S and since 1 ∈ S, 1 = sup(S). In this case,
sup(S) ∈ S.
Bounded below by −1 (since for n ∈ Z−{0}, either n ≥ 1, in which case
1
0 < n1 , or n ≤ −1, in which case n1 ≥ −1
= −1), and so has an infimum.
Since −1 is a lower bound for S and since −1 ∈ S, −1 = inf(S). In
this case, inf(S) ∈ S.
Since S is both bounded above and bounded below, it is bounded.
2. Bounded below by 0 (since 2x > 0 for all x ∈ R, we certainly have that
2x > 0 for all x ∈ Z), and so has an infimum. Given any ε > 0, we can
always find x so that 2x < ε, namely take log2 of both sides, and take
x to be any integer less than log2 (ε). Hence, there is no positive lower
bound, and so the greatest lower bound, the infimum, is inf(S) = 0.
Since there are no solutions to 2x = 0, in this case inf(S) 6∈ S.
1
Since 2x > x for positive integers x, given any C > 0 we can find an
x so that 2x > C, and so there is no upper bound. That is, S is not
bounded above.
Since S is not bounded above, it is not bounded.
3. Bounded below by −1 (since [−1, 1] = {x ∈ R | − 1 ≤ x ≤ 1} and since
−1 < 5), and so has an infimum. Since −1 is a lower bound for S and
since −1 ∈ S, −1 = inf(S). In this case, inf(S) ∈ S.
Bounded above by 5, and so has a supremum. Since 5 is an upper
bound for S and since 5 ∈ S, 5 = sup(S). In this case, sup(S) ∈ S.
Since S is both bounded above and bounded below, it is bounded.
4. Considering the subset of S in which y = 1, we have that S contains
the natural numbers N, and hence S is not bounded above.
Since x and 2y are both positive for x, y ∈ N, we have that 2xy > 0 for all
x, y ∈ N. Therefore, S is bounded below by 0, and so has an infimum.
Considering the subset of S in which x = 1, we have that S contains
1
for all y ∈ N. In particular, for each ε > 0, we can find y ∈ N so
2y
that 21y < ε, namely take log2 of both sides to get −y < log2 (ε), or
equivalently y > log2 (ε). Hence, there is no positive lower bound, and
so 0 = inf(S). Since 2xy is never 0 for x, y > 0, in this case inf(S) 6∈ S.
Since S is not bounded above, it is not bounded.
= 1 + n1 . Bounded below by 1, since n1 > 0 for all n ∈ N, and
5. Write n+1
n
hence 1 + n1 > 1 for all n ∈ N. Moreover, since for each ε > 1 we can
find n so that 1 + n1 < ε, there is no lower bound greater than 1, and
so inf(S) = 1. In this case, inf(S) 6∈ S, since 1 + n1 6= 1 for all n ∈ N.
Bounded above by 2, since n1 ≤ 1 for all n ∈ N and hence 1 + n1 ≤ 2.
In this case, 2 = 1 + 11 and so 2 ∈ S. Since 2 is an upper bound for S
that is contained in S, we have that 2 = sup(S) and so sup(S) ∈ S.
Since S is both bounded above and bounded below, it is bounded.
6. Break S up into two subsets, one of the positive terms (when n is even)
and the negative terms (when n is odd). So, S = {−2, − 34 , − 56 , . . .} ∪
{ 32 , 45 , 76 , . . .}.
The positive terms are all of the form 1 + n1 where n is even. Since n1
decreases as n increases, the largest positive term is 1 + 12 = 32 , and so
2
S is bounded above and hence has a supremum. Since S is bounded
above by 32 and since 23 ∈ S, sup(S) = 32 , and in this case sup(S) ∈ S.
The negative terms are all of the form 1 + n1 where n is odd. Since
1
decreases as n increases, − n1 increases as n increases, and so the
n
smallest negative term is −1−1
= −2, and so S is bounded below and
1
hence has an infimum. Since S is bounded below by −2 and since
−2 ∈ S, inf(S) = −2, and in this case inf(S) ∈ S.
Since S is both bounded above and bounded below, it is bounded.
√ √
7. We√can √rewrite S as S = (− 10, 10)
√ ∩ Q. By the definition of
(− 10, 10), S is bounded below by − 10, and hence
√ has an infimum.
Since there√are rational numbers greater than − 10 but arbitrarily
close
√ to − 10 (as can be seen by taking the decimal expansion of
− 10 and truncating
it after some number of places to get a√rational
√
number near −
10),
there
is no lower bound greater than − 10, and
√
so inf(S) = − 10. In this case, inf(S) 6∈ S.
√
S is bounded above by 10, and√hence has a supremum. Since
√ there
are rational numbers less than 10 but arbitrarily
√ close to 10 (as
can be seen by taking the decimal expansion of 10 and truncating
it
√
after some number of places to
√ get a rational number√near 10), there
is no upper bound less than 10, and so sup(S) = 10. In this case,
sup(S) 6∈ S.
Since S is both bounded above and bounded below, it is bounded.
8. Rewrite S as S = (−∞, −2)∪(2, ∞). This set is neither bounded above
(since for each real number r, there is s ∈ S with s > r, namely the
larger of 3 and r + 1) nor bounded below (since for each real number
r, there is s ∈ S with s < r, namely the smaller of −3 and r − 1).
Since S is not bounded below, it has no infimum. Since S is not
bounded above, it has no supremum.
Since S is neither bounded above not bounded below, it is not bounded.
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